Chapter 5.2, Problem 2E

During a recent drought, a water utility in a certain town sampled 100 residential water bills and found that 73 of the residences had reduced their water consumption over that of the previous year.

1. a. Find a 95% confidence interval for the proportion of residences that reduced their water consumption.
2. b. Find a 99% confidence interval for the proportion of residences that reduced their water consumption.
3. c. Find the sample size needed for a 95% confidence interval to specify the proportion to within ±0.05.
4. d. Find the sample size needed for a 99% confidence interval to specify the proportion to within ±0.05.
5. e. Someone claims that more than 70% of residences reduced their water consumption. With what level of confidence can this statement be made?
6. f. If 95% confidence intervals are computed for 200 towns, what is the probability that more than 192 of the confidence intervals cover the true proportions?

To determine

Find the 95% confidence interval for the proportion of residences that reduced their water consumption.

Answer to Problem 2E

The 95% confidence interval for the proportion of residences that reduced their water consumption is $\underset{\_}{\left(0.635,0.807\right)}$.

Explanation of Solution

Given info:

$X=73$ and $n=100$

Calculation:

Proportion:

The value of proportion is,

$\begin{array}{c}\tilde{p}=\frac{X+2}{\tilde{n}}\\ =\frac{X+2}{n+4}\\ =\frac{73+2}{100+4}\\ =0.7212\end{array}$

Confidence interval:

Step-by-step procedure to obtain the confidence interval using the MINITAB software:

• Choose Stat > Basic Statistics > 1 Proportion.
• Choose Summarized data.
• In Number of events, enter 75. In Number of trials, enter 104.
• Check Options; enter Confidence level as 95%.
• Choose not equal in alternative.
• Click OK.

Output using the MINITAB software is given below:

From the MINITAB output, the 95% confidence interval for the proportion of residences that reduced their water consumption is $\underset{\_}{\left(0.635,0.807\right)}$.

To determine

Find the 99% confidence interval for the proportion of residences that reduced their water consumption.

Answer to Problem 2E

The 99% confidence interval for the proportion of residences that reduced their water consumption is $\underset{\_}{\left(0.608,0.834\right)}$.

Explanation of Solution

Calculation:

Confidence interval:

Step-by-step procedure to obtain the confidence interval using the MINITAB software:

• Choose Stat > Basic Statistics > 1 Proportion.
• Choose Summarized data.
• In Number of events, enter 75. In Number of trials, enter 104.
• Check Options; enter Confidence level as 99%.
• Choose not equal in alternative.
• Click OK.

Output using the MINITAB software is given below:

From the MINITAB output, the 99% confidence interval for the proportion of residences that reduced their water consumption is $\underset{\_}{\left(0.608,0.834\right)}$.

To determine

Find the sample size needed for a 95% confidence interval to specify the proportion to within $\pm 0.05$.

Answer to Problem 2E

The sample size needed for a 95% confidence interval to specify the proportion to within $\pm 0.05$ is 305.

Explanation of Solution

Calculation:

Sample size:

The width for the 95% confidence interval is $\pm 1.96\sqrt{\frac{\tilde{p}\left(1-\tilde{p}\right)}{\left(n+4\right)}}$.

The width for the 95% confidence interval is for sample size n is $\pm 1.96\sqrt{\frac{0.7212\left(1-0.7212\right)}{\left(n+4\right)}}$ or $\pm \frac{0.8789}{\sqrt{\left(n+4\right)}}$.

By solving the equation,

$\begin{array}{c}\frac{0.8789}{\sqrt{\left(n+4\right)}}=0.05\\ \frac{0.8789}{0.05}=\sqrt{\left(n+4\right)}\\ {\left(\text{17}\text{.578}\right)}^2={\left(\sqrt{\left(n+4\right)}\right)}^2\\ \text{308}\text{.9861}=n+4\end{array}$

$\begin{array}{c}\text{308}\text{.9861}-4=n\\ \text{304}\text{.9861}=n\\ 305\approx n\end{array}$

Thus, the sample size needed for a 95% confidence interval to specify the proportion to within $\pm 0.05$ is 305.

To determine

Find the sample size needed for a 99% confidence interval to specify the proportion to within $\pm 0.05$.

Answer to Problem 2E

The sample size needed for a 99% confidence interval to specify the proportion to within $\pm 0.05$ is 532.

Explanation of Solution

Calculation:

Sample size:

The width for the 99% confidence interval is $\pm 2.58\sqrt{\frac{\tilde{p}\left(1-\tilde{p}\right)}{\left(n+4\right)}}$.

The width for the 99% confidence interval is for sample size n is $\pm 2.58\sqrt{\frac{0.7212\left(1-0.7212\right)}{\left(n+4\right)}}$ or $\pm \frac{1.1569}{\sqrt{\left(n+4\right)}}$.

By solving the equation,

$\begin{array}{c}\frac{1.1569}{\sqrt{\left(n+4\right)}}=0.05\\ \frac{1.1569}{0.05}=\sqrt{\left(n+4\right)}\\ {\left(\text{23}\text{.138}\right)}^2={\left(\sqrt{\left(n+4\right)}\right)}^2\\ \text{535}\text{.3670}=n+4\end{array}$

$\begin{array}{c}\text{535}\text{.3670}-4=n\\ \text{531}\text{.3670}=n\\ 532\approx n\end{array}$

Thus, the sample size needed for a 99% confidence interval to specify the proportion to within $\pm 0.05$ is 532.

To determine

Find the level of confidence.

Answer to Problem 2E

The level of confidence is 68.54%.

Explanation of Solution

Given info:

Here claim is that more than 70% of residences reduced their water consumption.

Calculation:

The value of $\sqrt{\frac{\tilde{p}\left(1-\tilde{p}\right)}{\left(n+4\right)}}$ is,

$\begin{array}{c}\sqrt{\frac{\tilde{p}\left(1-\tilde{p}\right)}{\left(n+4\right)}}=\sqrt{\frac{0.7212\left(1-0.7212\right)}{\left(100+4\right)}}\\ =0.0439\end{array}$

Confidence level:

Software Procedure:

Step-by-step procedure to obtain the confidence level using the MINITAB software:

• Choose Graph > Probability Distribution Plot choose View Probability> OK.
• From Distribution, choose ‘Normal’ distribution.
• Enter 0.7212 under Mean and Enter 0.0439 under Standard deviation.
• Click the Shaded Area tab.
• Choose X Value and Right Tail for the region of the curve to shade.
• Enter the X value as 0.7.
• Click OK.

Output using the MINITAB software is given below:

From the MINITAB output, the level of confidence is 0.6854 or 68.54%.

To determine

Find the probability that more than 192 of the confidence intervals cover the true proportions.

Answer to Problem 2E

The probability that more than 192 of the confidence intervals cover the true proportions is 0.2090.

Explanation of Solution

Given info:

$n=200$

Calculation:

Here X be the 95% confidence intervals cover the true proportions. Thus, $X\sim \text{Bin}\left(200,0.95\right)$.

Hence, $X\sim N\left(200\left(0.95\right),\sqrt{200\left(0.95\right)\left(0.05\right)}\right)$. That is, $X\sim N\left(190,3.0822\right)$

The probability that more than 192 of the confidence intervals cover the true proportions by using continuity correction is,

$\begin{array}{c}P\left(X>192\right)=P\left(X>192.5\right)\\ =P\left(z>\frac{192.5-190}{3.0822}\right)\\ =P\left(z>0.81\right)\\ =1-P\left(z<0.81\right)\end{array}$

$\begin{array}{c}=1-0.7910\\ =0.2090\end{array}$

Thus, the probability that more than 192 of the confidence intervals cover the true proportions is 0.2090.